Greg and Liz are currently 350 miles apart. They begin driving toward each other, Greg driving 60 mph and Liz driving 40 mph. How long will it take until they meet?
101
Answer: 3.5 hours
When dealing with two objects moving toward each other, we can simplify the calculations by adding the rates together. If Greg is moving 60 mph and Liz is moving 40 mph, then together they are traveling 100 miles per hour relative to one another. At that speed, they will collectively travel 350 miles in 3.5 hours, because T = (350 miles)/(100 mph) = 3.5.
Which is the correct expression for this problem?
8 students have been chosen to play for PCU's inter-collegiate basketball team. If every person on the team has an equal chance of starting, what is the probability that both Tom and Alex will start? (Assume 5 starting positions)
102
Answer: (A) 
First, count all the combinations of 5 people starting, given 8 people on the team. The formula is 
Next, to count the combinations that include Tom and Alex, assume that these 2 players start, then choose the remaining 3 starters from among the 6 remaining people. The formula is the same but takes different numbers: 
Finally, divide the second count by the first count, since Probability 
(as long as each combination is equally possible).
5 people in a company earn an average salary of $50,000. If 2 of the employees earn an average of $65,000, what is the average salary of the remaining 3 employees?
103
Answer: $40,000
One approach to this problem is to balance the “overs” and the “unders.” The 2 employees making $65,000 each make $15,000 more than the mean, for a total of $30,000 over the mean. That means the remaining 3 employees need to make a combined $30,000 under the mean. Distribute that amount evenly, and the remaining salaries average out to $10,000 below the mean, or $40,000.
10 years ago Tina was half as old as Ron will be 7 years from now. Which equation correctly represents this relationship?
(A) (1/2)(T – 10) = (R + 7)
(B) (T – 10) = (1/2)(R + 7)
104
Answer: (B) (T – 10) = (1/2)(R + 7)
When you translate sentences into equations, one of the easiest mistakes to make is to put the multiplier in the wrong place. Take the time to verify that it is where it should be. In this case, if Tina's age 10 years ago was half what Ron's age will be in 7 years, we need to multiply Ron's age by 1/2 to make them equal.
Don't forget to use the parentheses!
Jerry and Ross decide to have a footrace. They run 1,000 meters. Jerry runs 5 meters per second, and Ross runs 4 meters per second. Halfway through the race, Jerry realizes he is ahead and stops running for one full minute before finishing the race at his original speed. Who wins the race?
105
Answer: Ross wins
First, calculate how long it will take Ross to finish the race. T = D/R. T = (1,000 m)/(4m/s) = 250 s. To make the calculations simpler for Jerry, add 60 seconds to the total time to take into account the minute he spent not running. T = (1,000 m)/(5 m/s) + 60 s = 260 s. Ross finishes the race in less time.
What is the average of 3,456, 3,463, 3,470, 3,477, and 3,484?
106
Answer: 3,470
When a list is composed of evenly spaced numbers, the average will equal the median. In this case, all the numbers in the list are spaced 7 units apart. The median is 3,470, and the average is also 3,470.
The original ratio of girls to boys in a class was 2 : 3. Then 6 girls were added to the class, bringing the ratio of girls to boys in the class to 1 : 1. How many students are now in the class?
107
Answer: 36
Begin with the unknown multiplier. If the original ratio of boys to girls is 2 : 3, then the original number of girls is 2x and the original number of boys is 3x. After 6 girls are added, the new ratio is 1:1, so we can write the equation 
. Cross-multiply to get 2x + 6 = 3x, which simplifies to x = 6. So the original number of girls was 2x = 2(6) = 12 and the original number of boys was 3x = 3(6) = 18. There were 12 + 18 = 30 students in the class before the 6 girls were added. There are now 30 + 6 = 36 students in the class.
Car A travels east at 60 mph. Car B is 45 miles behind Car A and also travels east at 75 mph. How long will it take Car B to catch up with Car A?
108
Answer: 3 hours
When solving a combined rates problem with 2 objects moving in the same direction, we can ignore their actual speeds and focus only on the difference between their speeds. Car B is going 15 mph faster than Car A. That means it will catch up to Car A at a rate of 15 mph. Car B is currently 45 miles behind Car A, and R × T = D, so
(15 mph) × T = 45 miles. Solving for T gives us T = 3 hours.
A shipping company charges 5 + 10/y2 dollars per package shipped by a customer over a given month, where y is the number of packages shipped that month by the customer. If a customer spends $51 one month on shipping, which equation will correctly solve for the number of packages shipped that month by the customer?
(A) 5y + 10 = 51
(B) 5 + 10/y = 51
(C) 5y + 10/y = 51
109
Answer: (C) 5y + 10/y = 51
We know that y represents the number of packages shipped by the customer in the month, and 5 + 10/y2 is the shipping cost per package for that customer. Therefore, the product of these two terms must equal 51:
(y)(5 + 10/y2) = 51
5y + 10/y = 51
For added practice, what is the value of y? Try choosing integers that make sense.
Jeff can build a doghouse in 6 hours. Kevin can build the same doghouse in 3 hours. How long will it take them, working together, to build 1 doghouse?
110
Answer: 2 hours
Whenever you are told how long it takes a person to complete a task, a great first step is to turn that information into a rate. In an R × T = W equation, we can think of completing the task as doing 1 unit of work. If it takes Jeff 6 hours to build the doghouse, then his rate is 1/6 of the doghouse per hour. Similarly, Kevin's rate of work is 1/3 of the doghouse per hour. When we combine their rates (because they're working together), we see that they complete 1/2 of the doghouse every hour, because 1/6 + 1/3 = 1/2. At that rate, they will complete the doghouse in 2 hours.
In the World's Strongest Man competition, Olav Gundersson managed to pull a truck a total distance of 85 ft combined in two tries. On his second try, he pulled the truck 10 ft more than half the distance he pulled the truck on his first try. How far did he pull the truck on his first try?
(A) 45 ft (B) 50 ft (C) 55 ft (D) 60 ft (E) 65 ft
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Answer: (B) 50 ft
We can set this problem up and solve algebraically. We know that the sum of the two tries equals 85 feet, and the second try is 10 more than half the first try. Let's use x to represent the first try and y to represent the second try:
x + (10 + 0.5x) = 85
y = 10 + 0.5x
x + (10 + 0.5x) = 85
1.5x = 75
x = 50
Alternatively, plug the answer choices and eliminate.
John and Fawn, each drinking at a constant pace, can finish a case of soda together in 12 hours. If each were drinking the soda alone, it would take John 10 hours longer to finish the case of soda than it would take Fawn. In how many hours can John finish a case of soda, drinking alone?
(A) 24 hours (B) 30 hours (C) 36 hours (D) 42 hours
112
Answer: 30 hours
For all work problems, 
Here, the “work” is drinking one case of soda, so W = 1.
This problem provides three equations about rates: 
and 
and 
where J is John's rate, F is Fawn's rate, and t is John's time.
To avoid ugly algebra, work backwards from answer choices. Say we start with (B). John's time t = 30, so his rate J is 1/30. Fawn's time is t – 10 = 20, so her rate F is 1/20. Now check whether 1/30 + 1/20 = 1/12. This is true (1/30 + 1/20 = 2/60 + 3/60 = 5/60 = 1/12), so (B) is the answer.
A bag contains ping pong balls, each with a number written on it. The average of all the numbers is 50. When 
of the ping pong balls are removed, the average of the numbers written on the removed balls is 20. What is the average of the numbers on the balls still remaining in the bag?
113
Answer: 65
Although we don't know how many ping pong balls are in the bag, we can modify the weighted averages formula to help us answer this question. The weighted average equals the sum of the weights times the data points, divided by the sum of the weights. So, 
, or 
, and finally 
Alternatively, imagine that we start with 3 balls. Remove 1/3 of 3, or 1 ball, which has a 20 on it. The average is 50, so the sum is 3 × 50 = 150. The other two balls must add to 150 – 20 = 130. 130 averaged over 2 balls is 65.
At a farmer's market, one stall is selling mangos, kumquats and rutabagas. The ratio of mangos to kumquats is 5 : 4, and the ratio of kumquats to rutabagas is 3 : 7. Which of the following could be the number of kumquats?
(A) 28 (B) 36 (C) 42 (D) 49
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Answer: (B) 36
A common hidden constraint of word problems involving ratios is that the actual number of items must be an integer. For instance, it would not make sense in this context to have 5 1/3 mangos. The correct answer will be a multiple of both 4 and 3, based on the ratios involving kumquats.
For added practice, if there are actually 36 kumquats, how many mangos and rutabagas are there?
Train A and Train B start from 500 miles apart and travel towards each other on identical parallel tracks at constant speeds. They meet in 5 hours, but if Train B were going twice as fast as it actually is, the two trains would meet in 4 hours. At what rate is Train A actually traveling?
115
Answer: 75 mph
First, label the speed of Train A as RA and the speed of Train B as RB. When dealing with objects moving towards each other, we can add their rates, so (RA + RB) × (5 hrs) = 500 miles. The problem also states that (RA + 2RB) × (4 hrs) = 500 miles.
Dividing by the times:
RA + RB = 100 mph
RA + 2RB = 125 mph
Subtracting the first equation from the second:
RB = 25 mph
Therefore, RA = 75 mph.
What is the perimeter of Triangle BCD?
116
Answer: 
Because ∠ABD = ∠DAB, side BD has length 8. Because ∠BDA is 120°, ∠BDC is 60°, so triangle BCD is a 30–60–90 triangle.
| Side | Ratio | Length |
| CD | x | 4 |
| BC |  |
 |
| BD | 2x | 8 |
Sum the sides to find the perimeter: 8 + 4 + = 
If the area of triangle PQR is 30, and PR is the longest side of the triangle, what is the measure of ∠PQR?
117
Answer: 90°
Notice that ½(5)(12) = 30. QR is the base of the triangle and PQ is the height. Base and height of a triangle must be perpendicular to each other. Therefore ∠PQR is a right angle.
If PQR were anything other than a right triangle, the height of the triangle would be less than 12, and the area would be less than 30.
What is the area of a circle that has a circumference of 14π?
118
Answer: 49π
The key to this problem is to find the radius, because the radius is used to calculate both circumference and area.
C = 2πr, so 14π = 2πr. Therefore r = 7.
A = πr², so A = π(7)² = 49π
The ratio of side PR to side PQ is 5 : 4. What is the length of side QR?
119
Answer: 6
First, complete the ratio. 5/4 = 10/PQ. Side PQ has length 8. With two sides of a right triangle, you can use the Pythagorean theorem to find the third side. In this case, however, you should recognize that this triangle is a common Pythagorean triplet—a 6–8–10 triangle. Side QR has length 6.
A right circular cylinder has a volume of 10π. If the radius is doubled and the height remains the same, what is the volume of the new cylinder?
120
Answer: 40π
V = πr²h, so 10π = πr²h. We are told the radius is doubled, so the new radius is 2r. Replace r with 2r and the new equation is V = π(2r)²h. The new volume is V = 4 πr²h. Since we know that πr²h = 10π, we know that V = 4(10π) = 40π.
Alternately, try picking numbers for the cylinder that would produce an initial volume of 10π; a height of 10 and a radius of 1 are simplest. Double the radius and calculate the new volume.
What is the surface area of a cube with side length 5?
121
Answer: 150
Surface area is the sum of all the areas of the faces of the cube. The faces of a cube are all identical, so the surface area is the area of one of the faces times 6, because a cube has 6 faces. The area of one face is A = length × width.
Therefore, surface area = 6 × 5 × 5 = 150.
If PQ = QR, what is the area of triangle PQR?
122
Answer: 
Triangle OQR is a right triangle, and we have 2 sides, so either by using the Pythagorean Theorem or by recognizing it as a 30–60–90 triangle, we can find that the length of OQ is 
, which is also the height of triangle PQR. If PQ = QR, then ∠OPQ = ∠ORQ. Additionally, we know that ∠POQ = ∠ROQ = 90°. Therefore ∠OQP = ∠OQR, because when two triangles have two angles in common, they must have common third angles. Triangles OQR and OQP are thus similar triangles, and also have identical side lengths. OP must equal 3, so we have the length of the base as well as the height. The area of triangle PQR is 
If ∠PQR is a right angle, what is the area of the circle?
123
Answer: 
π or 6.25π
If ∠PQR is a right angle, then segment PR must be a diameter of the circle. We can solve for the length of PR using the Pythagorean Theorem, or by recognizing that triangle PQR is a 3–4–5 triangle. If the diameter of the circle is 5, then the radius is 2.5. The area of the circle is 
O is the center of the circle. The area of the circle is 81π. What is the length of line segment PR?
124
Answer: 
If the area of the circle is 81π, then 81π = πr², so r = 9. OP and OR are both radii, and so both have length 9. If triangle OPR is a right triangle and has 2 sides with equal length, then it is a 45–45–90 triangle, and the ratio of the sides to the hypotenuse is 
Therefore, the length of PR is 
If x = 70, what is y?
125
Answer: Maybe 110…maybe not!
Don't trust the picture! Although lines l and m appear to be parallel, nothing in the question tells us that they are. Unless explicitly told that the lines are parallel, there is no way to determine the value of y.
What is x?
126
Answer: 110°
The sum of the interior angles of a polygon = (n – 2) × 180, where n is the number of sides.
(6 – 2) × 180 = 720. Therefore, 5x + 170 = 720.
5x = 550, so x = 110.
Segment AD has length 15. What is the area of rectangle BCDE?
127
Answer: 120
If BCDE is a rectangle, then BC = ED. ED must have a length of 10, so AE has a length of 5. We can use the Pythagorean Theorem to determine the length of BE, or recognize that triangle ABE is a 5–12–13 triangle. Segment BE has a length of 12. Area of a rectangle is A = base × height = (10) × (12) = 120.
An empty right circular cylindrical swimming pool has a height of 10 ft; its base has an area of 15 ft². If water fills the pool at a rate of 25 ft³ every 10 minutes, how long will it take for the pool to be filled?
128
Answer: 60 Minutes
V = πr²h. Therefore, V = (15 ft²)(10 ft) = 150 ft³. The pool fills at a rate of 25 ft³ every 10 minutes, and since 25 × 6 = 150, we can proportionally apply this to the rate: 10 × 6 = 60 minutes.
What is the length of the main diagonal of a rectangular solid with sides of length 4, 4 and 2?
129
Answer: 6
We could use the Pythagorean Theorem twice, or we can save some time by using the “Deluxe” Pythagorean Theorem to find the interior diagonal of a rectangular prism (a box): d² = x² + y² + z².
So d² = 4² + 4² + 2² = 16 + 16 + 4 = 36, and d = 6.
If line l also passes through point (15, 6), what is the slope of line l?
130
Answer: 
Although we cannot easily see the point (15, 6) on the grid given in this problem, we know that two points define a line. Because line l goes through point (0, 3) and point (15, 6), we can calculate the slope. The slope is
The diagonal of a rectangular flowerbed is 
feet. What is the area of the flowerbed, if its length and width are equal?
131
Answer: 225 ft2
If the length and width are equal, then the flowerbed is a square, and the diagonal is the hypotenuse of a 45–45–90 triangle. The ratio of the sides to the hypotenuse is 
If the hypotenuse is 15
, then the sides each have length 15, so the area is l × w = (15) × (15) = 225.
O is the center of the circle. If minor arc PR has a length of π, what is ∠POR?
132
Answer: 60°
To determine ∠POR, we first need to determine what fraction of the circumference minor arc PR is. Circumference is C = 2πr, so C = 2π(3) = 6π. π/6π = 1/6, so minor arc PR is 1/6 of the circumference. That means ∠POR is 1/6 of 360°. ∠POR is 60°.
O is the center of the circle. What is ∠PRQ?
133
Answer: 30°
First, we can calculate ∠OPQ: since OP and OQ are the radii of the circle, they must have equal length, which means the opposite angles in triangle PQO must be equal. Therefore ∠OPQ = ∠PQO = 60°.
Since PR is a diameter of the circle, triangle PQR must be a right triangle. That means ∠PQR is a right angle, and ∠OQR is 90° – 60° = 30°.
Lastly, the three angles of triangle PRQ must add up to 180. ∠PRQ = 180° – 90° – 60° = 30°.
What is the distance between points (0,0) and (4,3)?
134
Answer: 5
One way to think of the distance formula is that it's the square root of the Pythagorean Theorem. The hypotenuse is the distance, the difference in the x-coordinates is the horizontal leg of the triangle, and the difference in the y-coordinates is the vertical leg. In this case, the two legs of the triangle have lengths 4 and 3, respectively. This is a 3–4–5 triangle in disguise. The hypotenuse of the triangle is 5.
What is the length of side SP?
135
Answer: 36
Although we can use the Pythagorean theorem, the numbers are large, and the calculation will be time-consuming. Remember that multiples of Pythagorean triplets are also Pythagorean triplets. This is a 3–4–5 triangle in disguise. Every value has been multiplied by 9. 3 × 9 = 27 and 5 × 9 = 45. The value we're missing is 4 × 9, which equals 36.
Look for common factors in side lengths such as these, to see whether the sides of the triangle are just multiples of a simpler Pythagorean triplet.
Which of these lines is perpendicular to the line whose equation is 
(A) 
(B) 
(C) 
136
Answer: (C) 
When determining what lines are perpendicular, there is only one important piece of information—the slope. The slopes of perpendicular lines are negative reciprocals. The slope of the original line is 2/3, so the slope of a perpendicular line will be –3/2. The slope of the line in answer choice (C) is –3/2.
If ∠POR = 58°, is point O the center of the circle?
137
Answer: No
∠PSR is an inscribed angle. If O were the center of the circle, ∠POR would be a central angle and angle ∠POR would be twice ∠PSR. Because 58° is not twice 30°, we know that point O cannot be the center of the circle.
Two sides of a triangle have lengths 3 and 9. Which of the following could be the length of the third side?
(A) 6 (B) 8 (C) 13
138
Answer: (B) 8
In any given triangle, the lengths of any two sides must add up to greater than the third side. Choice( A) fails, because 3 + 6 is not greater than 9. Choice (C) fails because 3 + 9 is not greater than 13.
Choice (B) works because:
3 + 6 > 8
3 + 8 > 6
6 + 8 > 3
In triangle ABC, AB = 12, AB = AC, and BD bisects AC. What is the area of triangle BDC?
139
Answer: 36
We know that AC has a length of 12, because it is equal to AB. We know that DC has a length of 6, because BD bisects AC. If DC is the base, then AB is the height of triangle BDC. A = ½(base) × (height), so A = ½(6) × (12) = 36.
The circumference of the smaller circle is 6π and the circumference of the larger circle is 16π. What is the area of the shaded region?
140
Answer: 55π
To find the area of the shaded region, we need to find the area of the entire figure and subtract the area of the smaller circle. The area of the entire figure is the area of the larger circle. If the larger circle has a circumference of 16π, then its radius is 8, because C = 2πr. The formula for area is A = πr², so A = π(8)² = 64π. Doing the same calculations for the smaller circle, we find it has a radius of 3 and an area of 9π. 64π – 9π = 55π.
O is the center of the circle and PS is a diameter of the circle. If the length of arc PQR is 4π, what is the length of line segment RS?
141
Answer: 
Since ∠ROS is a right angle, ∠POR is 180 – 90 = 90 degrees also, and 90° is 
of the circle. This implies that arc PQR is also of the circle. That means the circumference of the circle is 4 × 4π = 16π. Since C = 2πr = 16π, the radius r is 8.
In triangle ORS, both OR and OS are radii, so they form a right triangle with both legs = 8. ORS is a 45–45–90 triangle, in which the hypotenuse is 
times the leg length. Thus, line segment RS is in length.
If the area of square BCEF is 16 and the length of AD is 10, what is the area of trapezoid ADEF?
142
Answer: 28
The area of a trapezoid is 
× (height), where b1 and b2 are the parallel bases of the trapezoid. We can determine the height, because a square that has an area of 16 must have sides of length 4, and BF is the height of the trapezoid. From the statement, we know the average of the lengths of the parallel bases is 
, so the area is A = 7 × 4 = 28.
Segment BE bisects segment AD. What is the area of parallelogram ABCD?
143
Answer: 24
The area of a parallelogram is A = base × height. If BC has length 8, then AD also has length 8. If AD is the base, then BE can be the height, because base and height must be perpendicular. We are told BE bisects AD, so AE must have length 4. We see then that triangle ABE is a 3–4–5 triangle in disguise. BE has length 3, and the area is (8)(3) = 24.
| Quantity A | Quantity B |
| (3)(5) | 3 + 5 |
144
Answer: (A) Quantity A is greater.
Quantity A: (3)(5) = 15
Quantity B: 3 + 5 = 8
| Quantity A | Quantity B |
| 3(5 + 1) | (3)(5) + (3)(1) |
145
Answer: (C) The two quantities are equal.
Quantity A: 3(5 + 1) = 3(6) = 18
Quantity B: (3)(5) + (3)(1) = 15 + 3 = 18
Alternatively, note that Quantity B is the distributed form of Quantity A, or Quantity A is the factored form of Quantity B.
| Quantity A | Quantity B |
| 3(5 + 1) | (3)(5)(1) |
146
Answer: (A) Quantity A is greater.
Quantity A: 3(5 + 1) =3(6) = 18
Quantity B: (3)(5)(1) = 15
Or, note that both quantities are 3 times something, and that 5 +1 > 5.
| Quantity A | Quantity B |
| 3(5 + 2) |  |
147
Answer: (A) Quantity A is greater.
Quantity A: 3(5 + 2) = 3(7) = 21
Quantity B:
Or, note that both quantities are 3 times something. We can compare 7 to the reciprocal of 
, which will be less than than 7, since the fraction sum is greater than than 1/7.
| Quantity A | Quantity B |
| 47% of 1,200 | 153% of 400 |
148
Answer: (B) Quantity B is greater.
Quantity A: 47% of 1,200 is a bit less than half of 1,200. Less than 600.
Quantity B: 153% of 400 is a bit more than 1.5 times 400. More than 600.
| Quantity A | Quantity B |
| BC | CD |
149
Answer: (A) Quantity A is greater.
Drawing the figure to scale helps quite a bit:
BC is longer than CD by a significant margin; you can trust the to-scale picture.
AC splits right triangle ABD into two more similar right triangles: ABC and ACD. In similar triangles, the ratio of short leg to the long leg is constant (here, a ratio of 1:2). In ABC, BC is the long leg and AC is the short leg. In ACD, AC is the long leg and CD is the short leg. So, BC > AC > CD.
| y = 3x2 – 14x + 5 | |
| Quantity A | Quantity B |
| x | y |
150
Answer: (D) The relationship cannot be determined from the information given.
Try:
If x = 0, y = 0 – 0 + 5 = 5. Since 5 > 0, y > x.
If x = 1, y = 3 – 14 + 5 = –6. Since 1 > –6, x > y.